Topological Completeness is Weakly Hereditary
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Theorem
Let $T = \struct {S, \tau}$ be a topological space which is topologically complete.
Let $V \subseteq S$ be a closed subspace of $T$.
Then $V$ is also topologically complete.
That is, topological completeness is weakly hereditary.
Proof
This needs considerable tedious hard slog to complete it. In particular: Use Subspace of Complete Metric Space is Closed iff Complete To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Finish}} from the code.If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page. |
Sources
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $5$: Metric Spaces: Complete Metric Spaces